# Degree of splitting fields over finite fields

Suppose I have a finite field $$F_{p^d}$$, and I have a polynomial $$f$$ which is of degree $$n$$ and irreducible. I have a feeling that the splitting field of $$f$$ over $$F_{p^d}$$ is $$F_{p^{dn}}$$, but I am not sure how to show this.

I think that since finite fields are perfect, every irreducible polynomial is separable, meaning that the splitting field of $$f$$ is $$F_{p^d}(\alpha)$$ where the degree of the minimal polynomial of $$\alpha$$ is $$n$$, and so the splitting field is like a degree $$n$$ vector space over $$F_{p^d}$$.

• For all $m$ and for all agebraically closed fields $k$ of characteristic $p$ there is exactly one subfield $F\subseteq k$ such that $\lvert F\rvert=p^m$. Namely, it can be characterized as the splitting field of the polynomial $x^{p^m}-x$ over the base field. For degree reasons, every root $\alpha$ of $f$ must satisfy $F_{p^d}(\alpha)=F_{p^{dn}}$. – Saucy O'Path May 8 at 0:07